Final

Problem 1

Introduction

Methods

Loading Data

# loading data
kang_seed <- read.table(here("data/sev208_kratseedbank_20120213.txt"), sep=',', header = TRUE) |> 
             mutate(mnd = as.factor(mnd)) |> 
             filter(!(species %in% c("soil", "plant", "dist", "litter", "gravel")))

Missing Data

gg_miss_var(kang_seed)

Skim

skim(kang_seed)

Data summary
Name	kang_seed
Number of rows	960
Number of columns	5
_______________________
Column type frequency:
character	3
factor	1
numeric	1
________________________
Group variables	None

Variable type: character

skim_variable	complete_rate	min	max	n_unique
dir	1	1	1	4
loc	1	1	1	4
species	1	4	6	8

Variable type: factor

skim_variable	n_missing	complete_rate	ordered	n_unique	top_counts
mnd	0	1	FALSE	10	18: 128, 23: 128, 25: 128, 6: 96

Variable type: numeric

skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
seeds	0	1	11.1	48.21	0	0	0	3	656	▇▁▁▁▁

Visualize Data

# visualize counts
ggplot(kang_seed, aes(x = seeds)) +
  geom_histogram(bins = 17)

#visualize using box plot
ggplot(data = kang_seed, aes(x = mnd, y = seeds)) +
  geom_boxplot() +
  labs(x = "Mound Number", y = "NUmber of Seeds", title = "Boxplot of Seeds Found at Different Mounds") +
  theme_bw()

# set up data for bar graph
kang_seed_bar <- kang_seed |> 
                 group_by(mnd) |> 
                 summarise(across(seeds, sum)) |> 
                 ungroup()
# bar graph
ggplot() +
  geom_bar(data = kang_seed_bar, aes(x = mnd, y = seeds, fill = mnd), 
           stat = "identity", show.legend = F) +
  theme_bw() +
  labs(x = "Mound Number", y = "Number of Seeds", title = "Total Seed Count per Mound")

Question: How does total seed number differ between kangaroo rat mound locations?

Choose Test

We will be performing an ANOVA test to determine if there is a difference in the total seed count between kangaroo rat mound locations.

Set Up Hypothesis

\(H_0:\) There is no significant difference in seed counts between the kangaroo rat mound locations. \(H_a:\) There is a significant difference in seed counts between at least two of the kangaroo mound locations.

Initial Assumption Checks

# make aov object
kang_aov <- aov(seeds ~ mnd, data = kang_seed)

# get resiudals
kang_res <- kang_aov$residuals

# check normality
qqPlot(kang_res)

## [1] 899 195

The data does not appear to be normal, which we expceted because in the context reading they discussed how in their analysis the data was non-normal even after trying transformations. We decided to try a log transformation regaudless because this would be the logical next step in this analysis.

# transform data
kang_seed_log <- kang_seed |> 
                 mutate(seeds = log(seeds + 1))

# make new aov object
kang_log_aov <- aov(seeds ~ mnd, data = kang_seed_log)

# residuals
kang_log_res <- kang_log_aov$residuals

#test normality
qqPlot(kang_log_res)

## [1] 899 195

The data is still non-normal as expected (we mentionded above that the context paper mentioned how the data is not normal)

# show aov summary
summary(kang_aov)

##              Df  Sum Sq Mean Sq F value Pr(>F)
## mnd           9   15897    1766   0.758  0.655
## Residuals   950 2212569    2329

The ANOVA p-value is greater than 0.05, meaning we fail to reject the null hypothesis that there is no significant difference in total seed count between different kangaroo rat mounds.

Results

Problem 2

Introduction

Methods

Loading Data

# loading data
shrub_raw <- read.csv(here("data/shrubstudy_seed.csv"))

shrub_seed <- shrub_raw |>
              dplyr::select(treatment, species, total_inf = total_nr_infl, 
                      num_seeds = nr_seeds, shrub_num, plant_num = plant_nr, tag_num) |>
              mutate(treatment = as.factor(treatment))

Missing Data

# visualize missing data
gg_miss_var(shrub_seed)

Skim

skim(shrub_seed)

Data summary
Name	shrub_seed
Number of rows	287
Number of columns	7
_______________________
Column type frequency:
character	1
factor	1
numeric	5
________________________
Group variables	None

Variable type: character

skim_variable	n_missing	complete_rate	min	max	empty	n_unique	whitespace
species	0	1	6	6	0	19	0

Variable type: factor

skim_variable	n_missing	complete_rate	ordered	n_unique	top_counts
treatment	0	1	FALSE	2	con: 174, shr: 113

Variable type: numeric

skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
total_inf	0	1.00	5.15	10.45	1	1.00	1	5.00	117	▇▁▁▁▁
num_seeds	105	0.63	14.55	28.62	0	1.25	5	13.75	285	▇▁▁▁▁
shrub_num	0	1.00	29.53	16.04	5	13.00	30	44.00	54	▆▁▇▂▅
plant_num	0	1.00	2.43	1.34	1	1.00	2	3.00	5	▇▆▅▃▂
tag_num	0	1.00	143.48	63.71	22	104.00	159	181.00	300	▃▂▇▂▁

Subset Dataset

# filter out missing data
shrub_seed_sub <- shrub_seed |> 
                  drop_na(num_seeds)

Histogram of Counts

ggplot(shrub_seed_sub, aes(x = num_seeds)) +
  geom_histogram(bins = 17)

Correlation Plot

# calc Pearson's r for numerical values
shrub_seed_sub_cor <- shrub_seed_sub |> 
                      dplyr::select(3:4) |> 
                      cor(method = "pearson")

# plot correlation values
corrplot(shrub_seed_sub_cor,
         method = "ellipse",
         addCoef.col = "black")

Variable Relationships

shrub_seed_sub |> 
  dplyr::select(!num_seeds) |> 
  ggpairs()

Question: How does seed count vary with plot type (shrub or open), plant species, and total number of inflorescences? Is there a simpler model that explains seed count, and if so, what is it?

Build Models

# linear model, we know this is wrong
shrub_model1 <- lm(num_seeds ~ treatment + species + total_inf, data = shrub_seed_sub)

# generalized linear model with Poisson distribution
shrub_model2 <- glm(num_seeds ~ treatment + species + total_inf, data = shrub_seed_sub, family = "poisson")

# generalized linear model with negative binomial distribution
shrub_model3 <- glmmTMB(num_seeds ~ treatment + species + total_inf, data = shrub_seed_sub, family = "nbinom2")

# generalized linear model with Poisson distribution and random effect of treatment
shrub_model4 <- glmmTMB(num_seeds ~ treatment + species + total_inf + (1|shrub_num), 
                          data = shrub_seed_sub, family = "poisson")

# generalized linear model with negative binomial distribution and random effect of treatment
shrub_model5 <- glmmTMB(num_seeds ~ treatment + species + total_inf + (1|shrub_num), 
                      data = shrub_seed_sub, family = "nbinom2")

Because we are looking at count data, we know that the data is discrete and only has a lower bound. Knowing this, we built a couple different models using the Poisson and Negative Binomial distribution.

# check diagnostics
simulationOutput_m1 <- simulateResiduals(shrub_model1)

plot(simulationOutput_m1)

## qu = 0.25, log(sigma) = -2.738927 : outer Newton did not converge fully.

## qu = 0.75, log(sigma) = -2.138039 : outer Newton did not converge fully.

par(mfrow = c(1,2))

testDispersion(simulationOutput_m1)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 0.95995, p-value = 0.64
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m1)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = Inf, p-value < 2.2e-16
## alternative hypothesis: two.sided

# check diagnostics
simulationOutput_m2 <- simulateResiduals(shrub_model2)

plot(simulationOutput_m2)

## qu = 0.5, log(sigma) = -2.296452 : outer Newton did not converge fully.

par(mfrow = c(1,2))

testDispersion(simulationOutput_m2)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 20.946, p-value < 2.2e-16
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m2)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = 3.8259, p-value < 2.2e-16
## alternative hypothesis: two.sided

simulationOutput_m3 <- simulateResiduals(shrub_model3)

plot(simulationOutput_m3)

par(mfrow = c(1,2))

testDispersion(simulationOutput_m3)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 1.6709, p-value = 0.208
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m3)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = 1.4543, p-value = 0.024
## alternative hypothesis: two.sided

# check diagnostics
simulationOutput_m4 <- simulateResiduals(shrub_model4)

plot(simulationOutput_m4)

par(mfrow = c(1,2))

testDispersion(simulationOutput_m4)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 0.80605, p-value = 0.576
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m4)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = 3.2154, p-value < 2.2e-16
## alternative hypothesis: two.sided

# check diagnostics
simulationOutput_m5 <- simulateResiduals(shrub_model5)

plot(simulationOutput_m5)

par(mfrow = c(1,2))

testDispersion(simulationOutput_m5)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 1.6618, p-value = 0.192
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m5)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = 1.4583, p-value = 0.04
## alternative hypothesis: two.sided

Based on the tests above, it appears that either shrub_model3 or shrub_model5 will be the best models because neither are over or under-dispered. However, they do still have some patterns in the residuals, and they appear to be zero-inflated. To further our model, we will create a new model with a zero-inflated term.

# generalized linear model with negative binomial distribution
shrub_model6 <- glmmTMB(num_seeds ~ treatment + species + total_inf, ziformula = ~1,
                        data = shrub_seed_sub, family = "nbinom2")

With the zero-inflated model created, we will now run the necessary diagnostics.

# check diagnostics
simulationOutput_m6 <- simulateResiduals(shrub_model6)

plot(simulationOutput_m6)

par(mfrow = c(1,2))

testDispersion(simulationOutput_m6)

## 
##  DHARMa nonparametric dispersion test via sd of residuals fitted vs.
##  simulated
## 
## data:  simulationOutput
## dispersion = 1.7383, p-value = 0.36
## alternative hypothesis: two.sided

testZeroInflation(simulationOutput_m6)

## 
##  DHARMa zero-inflation test via comparison to expected zeros with
##  simulation under H0 = fitted model
## 
## data:  simulationOutput
## ratioObsSim = 1.0537, p-value = 0.824
## alternative hypothesis: two.sided

Based on the above plots, shrub_model6 appears to be our best model. There is no over or underdispersion, and our distribution has about the same number of zeros as our model would predict. There is still some trend in our residuals, but it appears to have been lessened when compared to shrub_model3. We acknowledge this trend in residuals and have decided that isn’t a big cause for concern since the other diagnostics were met.

Moving forward, we will check AICC values to test whether our assumptions about the models are correct.

Which distribution to use?

MuMIn::model.sel(shrub_model1, shrub_model2, shrub_model3, shrub_model4, shrub_model5, shrub_model6)

## Model selection table 
##              (Intr) spc ttl_inf trt cnd((Int)) dsp((Int)) cnd(spc) cnd(ttl_inf)
## shrub_model6                             2.158          +        +      0.06625
## shrub_model3                             1.918          +        +      0.07499
## shrub_model5                             1.916          +        +      0.07501
## shrub_model1 -2.567   + 2.15000   +                                            
## shrub_model4                             2.488          +        +      0.03145
## shrub_model2  2.549   + 0.02989   +                                            
##              cnd(trt) zi((Int)) family   class random df    logLik   AICc
## shrub_model6        +    -2.281 n2(lg) glmmTMB        10  -553.169 1127.6
## shrub_model3        +           n2(lg) glmmTMB         9  -556.455 1132.0
## shrub_model5        +           n2(lg) glmmTMB c(s_n) 10  -556.282 1133.9
## shrub_model1                    gs(id)      lm         9  -692.468 1404.0
## shrub_model4        +           ps(lg) glmmTMB c(s_n)  9 -1058.768 2136.6
## shrub_model2                    ps(lg)     glm         8 -1150.107 2317.0
##                delta weight
## shrub_model6    0.00  0.863
## shrub_model3    4.33  0.099
## shrub_model5    6.23  0.038
## shrub_model1  276.36  0.000
## shrub_model4 1008.96  0.000
## shrub_model2 1189.42  0.000
## Abbreviations:
##  family: gs(id) = 'gaussian(identity)', n2(lg) = 'nbinom2(log)', 
##          ps(lg) = 'poisson(log)'
## Models ranked by AICc(x) 
## Random terms: 
##  c(s_n): cond(1 | shrub_num)

This chart confirms our assumptions about our models; shrub_model6 is best, followed by shrub_model3 and shrub_model5. We know this because of their respective AICC values of 1127.6, 1132.0, and 1133.9. In general, a lower AICC value equates to a better model.

Next, we will look closer at our model: shrub_model6.

Model Summary

# summary
summary(shrub_model6)

##  Family: nbinom2  ( log )
## Formula:          num_seeds ~ treatment + species + total_inf
## Zero inflation:             ~1
## Data: shrub_seed_sub
## 
##      AIC      BIC   logLik deviance df.resid 
##   1126.3   1158.4   -553.2   1106.3      172 
## 
## 
## Dispersion parameter for nbinom2 family (): 2.21 
## 
## Conditional model:
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept)     2.158294   0.211515  10.204  < 2e-16 ***
## treatmentshrub -0.298882   0.131165  -2.279   0.0227 *  
## speciesCARRUP  -1.720761   0.280803  -6.128  8.9e-10 ***
## speciesGEUROS  -0.239652   0.241272  -0.993   0.3206    
## speciesKOBMYO   0.045647   0.199786   0.228   0.8193    
## speciesMINOBT  -0.416863   0.210918  -1.976   0.0481 *  
## speciesTRIDAS   1.486501   0.724587   2.052   0.0402 *  
## total_inf       0.066248   0.007463   8.877  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Zero-inflation model:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -2.2813     0.3944  -5.784  7.3e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The summary of our model shows that treatmentshrub, speciesCARRUP, speciesMINOBT, speciesTRIDAS, and total_inf all appears to be significant indicators of total number of seeds.

# confidence intervals
confint(shrub_model6)

##                           2.5 %       97.5 %    Estimate
## cond.(Intercept)     1.74373255  2.572855217  2.15829389
## cond.treatmentshrub -0.55596148 -0.041802243 -0.29888186
## cond.speciesCARRUP  -2.27112456 -1.170397104 -1.72076083
## cond.speciesGEUROS  -0.71253699  0.233232924 -0.23965203
## cond.speciesKOBMYO  -0.34592667  0.437221033  0.04564718
## cond.speciesMINOBT  -0.83025460 -0.003471633 -0.41686312
## cond.speciesTRIDAS   0.06633737  2.906665530  1.48650145
## cond.total_inf       0.05162033  0.080875060  0.06624770
## zi.(Intercept)      -3.05433608 -1.508209420 -2.28127275

# adjusted R2
r.squaredGLMM(shrub_model6)

##                 R2m       R2c
## delta     0.7256489 0.7256489
## lognormal 0.7667371 0.7667371
## trigamma  0.6697461 0.6697461

Results

Table Format

# model object in table
shrub_model6 |> 
  as_flextable()

		Estimate	Standard Error	statistic	p-value
Fixed effects
	(Intercept)	2.158	0.212	10.204	0.0000	***
	treatmentshrub	-0.299	0.131	-2.279	0.0227	*
	speciesCARRUP	-1.721	0.281	-6.128	0.0000	***
	speciesGEUROS	-0.240	0.241	-0.993	0.3206
	speciesKOBMYO	0.046	0.200	0.228	0.8193
	speciesMINOBT	-0.417	0.211	-1.976	0.0481	*
	speciesTRIDAS	1.487	0.725	2.052	0.0402	*
	total_inf	0.066	0.007	8.877	0.0000	***
	(Intercept)	-2.281	0.394	-5.784	0.0000	***
Signif. codes: 0 <= '*' < 0.001 < '' < 0.01 < '*' < 0.05

square root of the estimated residual variance: 2.2
data's log-likelihood under the model: -553.2
Akaike Information Criterion: 1,126.3
Bayesian Information Criterion: 1,158.4

The above displays what we studied earlier in the summary, just in an easier to digest format.

Visualization

plot(allEffects(shrub_model6))

To answer the question we are studying, how does seed count vary with plot type (shrub or open), plant species, and total number of inflorescences, we need to look at the effects of each of the variables. This plot shows that the number of seeds is generally higher in the open (control) plot. Additionally, the species TRIDAS tends to have the highest number of seeds, whereas CARRUP appears to have the lowest amount of seeds. Lastly, the number of seeds increases as the number of total inflorescenses increases.

Final

Adam Guerra, Luke Todd, Riley Zamora

2023-06-11

Problem 1

Introduction

Methods

Loading Data

Missing Data

Skim

Visualize Data

Choose Test

Set Up Hypothesis

Initial Assumption Checks

Results

Problem 2

Introduction

Methods

Loading Data

Missing Data

Skim

Subset Dataset

Histogram of Counts

Correlation Plot

Variable Relationships

Build Models

Which distribution to use?

Model Summary

Results

Table Format

Visualization